Evaluation of Integrals

If   $F(s) = \mathcal{L}\left\{ f(t) \right\}$,   then   $\displaystyle \int_0^\infty e^{-st} f(t) \, dt = F(s)$.
 

Taking the limit as   $s \to 0$,   then   $\displaystyle \int_0^\infty f(t) \, dt = F(0)$   assuming the integral to be convergent.
 

Laplace Transform of Intergrals

Theorem
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then
 

$\displaystyle \mathcal{L} \left[ \int_0^t f(u) \, du \right] = \dfrac{F(s)}{s}$

 

Laplace Transform of Derivatives

For first-order derivative:
$\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$
 

For second-order derivative:
$\mathcal{L} \left\{ f''(t) \right\} = s^2 \mathcal{L} \left\{ f(t) \right\} - s \, f(0) - f'(0)$
 

For third-order derivative:
$\mathcal{L} \left\{ f'''(t) \right\} = s^3 \mathcal{L} \left\{ f(t) \right\} - s^2 f(0) - s \, f'(0) - f''(0)$
 

For nth order derivative:

$\mathcal{L} \left\{ f^n(t) \right\} = s^n \mathcal{L} \left\{ f(t) \right\} - s^{n - 1} f(0) - s^{n - 2} \, f'(0) - \dots - f^{n - 1}(0)$

 

Division by t | Laplace Transform

Division by $t$
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,
 

$\displaystyle \mathcal{L} \left\{ \dfrac{f(t)}{t} \right\} = \int_s^\infty F(u) \, du$

 

provided   $\displaystyle \lim_{t \rightarrow 0} \left[ \dfrac{f(t)}{t} \right]$   exists.
 

Second Shifting Property | Laplace Transform

Second Shifting Property
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   and   $g(t)
= \begin{cases} f(t - a) & t \gt a \\ 0 & t \lt a \end{cases}$
 

then,

$\mathcal{L} \left\{ g(t) \right\} = e^{-as} F(s)$